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3 years ago

Which planet is closest to the sun? Venus Pluto Neptune Mercury

2 answers:

6 0

<span>The closest planet to the sun is Mercury. After Mercury the order of the planets is Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune. </span>

arlik [135]3 years ago

3 0

<span>Mercury is the closest planet to the Sun. </span>

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In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

nata0808 [166]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

7 0

3 years ago

Most towns use a water tower to store water and provide pressure in the pipes that deliver water to customers. The figure below

MatroZZZ [7]

The effective height of the water for Smith's house will be 24.61m.

<h3>How to calculate the height?</h3>

Based on the information given, the volume of the water in sphere will be:

= 4/3πr³ = (5.80 × 10^5)/1000

= 4.18r³ = 580

r³ = 138.7

r = 5.18m

The effective height of the water will be:

= 18.0 + 2(5.18)

= 28.36

The gauge pressure at Faucet of Jones house will be:

= pgh

= 1000(9.8)(28.36)

= 277.9kPa

The effective height of the water for Smith's house will be:

= 18.0 + 2(5.18) - 3.75

= 24.61m

The gauge pressure at Faucet of Jones house will be:

= 1000 × 9.8 × 24.61

= 241.2kPa

Learn more about height on:

brainly.com/question/983412

#SPJ1

8 0

1 year ago

g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7

bogdanovich [222]

Answer:

When they are approaching each other

$f_a = 2228.7 \ Hz$

When they are passing each other

$f_a = 2100Hz$

When they are retreating from each other

$f_a = 1980.7 Hz$

Explanation:

From the question we are told that

The velocity of car one is $v_1 = 13.0 m/s$

The velocity of car two is $v_2 = 7.22 m/s$

The frequency of sound from car one is $f_e = 2.10 kHz$

Generally the speed of sound at normal temperature is $v = 343 m/s$

Now as the cars move relative to each other doppler effect is created and this can be represented mathematically as

$f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]$

Where $v_s$ is the velocity of the source of sound

$v_o$ is the velocity of the observer of the sound

$f_o$ is the actual frequence

$f_a$ is the apparent frequency

Considering the case when they are approaching each other

$f_a = f_o [\frac{v + v_o}{v - v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 + 7.22}{ 343 - 13} ]$

$f_a = 2228.7 \ Hz$

Considering the case when they are passing each other

At that instant

$v_o = v_s = 0m/s$

$f_o = f_e$

$f_a = f_o [\frac{v }{v } ]$

$f_a = f_o$

Substituting value

$f_a = 2100Hz$

Considering the case when they are retreating from each other

$f_a = f_o [\frac{v - v_o}{v + v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 - 7.22}{343 + 13} ]$

$f_a = 1980.7 Hz$

7 0

3 years ago

A skateboarder shoots off a ramp with a velocity of 5.1 m/s, directed at an angle of 55° above the horizontal. The end of the ra

ExtremeBDS [4]

Answer

given,

initial velocity of skateboard = 5.1 m/s

angle above the horizontal = 55°

height of the ramp = 1 m

a) maximum height of projectile

$H = \dfrac{u^2sin^2 \theta}{2g}$

$H = \dfrac{5.1^2\times sin^2 55^0}{2\times 9.81}$

H = 0.889 m

the maximum height of the skateboard above the ground

= 1 + 0.889

= 1.889 m

b) time to reach the height

$t = \dfrac{u\ sin\theta}{g}$

$t = \dfrac{5.1\ sin55^0}{9.8}$

t = 0.426 s

horizontal distance = u cos θ × t

= 5.1 × cos 55° × 0.426

horizontal distance = 1.25 m

7 0

3 years ago

Como estan formados los musculos?

VikaD [51]

Answer:

Explanation: El músculo esquelético está formado por fibras musculares, rodeadas de una capa de tejido conjuntivo, denominada endomisio. Las fibras se reúnen en fascículos primarios, que también están rodeados por otra capa de tejido conjuntivo, esta vez, más grueso, denominada perimisio.

3 0

3 years ago

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